PYQ on Uniqueness and Existence in ODE | Short Cut Tricks | CSIR NET 2011 to 2023

[@debajitroy7300]

one of "The" best channel for net,Gate,JAM

[@Harish-Dhar-Deewan]

Gajab ka fast trick btate ho sir.....step function wala shot trick aaj hi pta chla ......book me bhut lamba calcultion diya hai

[@udaybhansahu2971]

❤very nice explanation

[@SADDAMHUSSAIN-mw3cv]

Superb...

[@MohdAasif-mf8tu]

@11:22 e^z is not bounded. It's unbounded

[@shweta9227]

Yes

[@stevenabakash919]

Yes, entire nonconstant function is unbounded

[@divyaraju-y4k]

best channel i ever seen
thank u sir! pls continue more

[@mymathspartner6019]

Nice video sir, thankyou sir

[@ritutiwari4687]

Nice ❤

[@mathbasicbox5249]

❤❤ what a approach

[@simisaharan8754]

Nice explanation sir.

[@shivamupd]

It's very helpful video for us.

[@RanjangamingYT7477]

1:3:25 check y=y0 e^x
Then y0 non zero but y0 positive and negative ho sakta ha some time increase and some time decrease
Therefore, not monotonic

[@Harish-Dhar-Deewan]

Sahi kaha aapne

[@Harish-Dhar-Deewan]

We can't say increasing or decreasing function

[@theexpress85]

No words 🙏🙏🙏🙏🙏🙏🙏

[@justiceforsidhumoosewala6225]

Nice explanation sir ❤

[@shrawankumarpatel2148]

Thank you sir

[@ronneysaini7763]

Thanks Sir❤❤❤😊

[@triptikumari9020]

Thank you 👍👍 Sir

[@professorragavan9783]

Thank you sir!

[@Manish-uq5le]

Thank you sir❤

[@anshikathakur7085]

Such a great explanation thanku so much sir means a lot God bless uh with lot of success and more❤

[@Ayoknyising]

Sir you are alys great... ❤

[@yashodhas1174]

Thanks a lot sir, for this wonderful and informative class..😊

[@melodiousprabahini]

Thank you sir. So helpful vdo

[@simonrai7300]

Thank you Sir.
❤❤❤❤

[@easymathematics1341]

Ever greatest sir

[@mathbasicbox5249]

Thanku so much sir ur great sir

[@naveed6396]

At 7:01 what about the case when A0.

[@nishajyoti5896]

एक्सीलेंट 🎉

[@parveenkumar8050]

Thankyou sir

[@sibashishdas8806]

1:06:33 I think it's z=1/1+t

[@someshnogia2009]

Yes you are right.

[@avipsaswain1162]

Yes

[@shyamnagar8917]

Tnkyu sir ❤

[@subratpradhan3581]

🙏🙏🙏 Thanks sir
50:16

[@bakulmanna7123]

Thank you so much sir ❤❤

[@mamtakeswani3066]

sir please make videos for solutions of ring theory and group theory

[@subratpradhan3581]

Sir has no videos on Group and Ring theory, he has mastery in applied portions + Real Analysis+ CA + Linear Algebra

[@memebaaz4005]

Superhlb

[@anshukumartiwari1361]

Result @7:03 is incorrect for A

[@Seema-jf9nr]

Answer to last ques " blows up in finite time" is option B not D, sir plz recheck

[@kshamashreek6126]

Last question... i think the right answer is option b

[@meghavar]

Sir WhatsApp group join ni ho pata iss link se

[@vindhyavijayan3826]

Sir in 1st question
Since every polynomial is continuous atleast one solution will exists. How is possible for no solution?

[@Kawaionnanoko8777]

Maybe the roots lie in the complex plane

[@SasankaKayal-b8s]

Sir please making vidios for super tricks solution of algebra group theory and ring theory .....other vidio are very helpful ...so thank you so much sir ❤

[@ronneysaini7763]

Nahi banayenge sir

[@chinmoykumarpal5461]

9:25 here (u -1)^2 is continuous so by existence theorem solution exists...as you mention in that question 37:55
37:55 here if I take (y-1)^2 and y(0)= 1 then no solution exists which u have done in 9:25

[@poojasingh7200]

As you told in this vdo
e^x is unbounded function this is correct but
e^z is bounded function ....how can you say
We know that non constant entire function is unbounded
So
e^z is also unbounded function

[@Ekshivbhakat]

Thank you so much sir for helping us 🙏🙏

[@RoshanKumar-maths]

52:12 answer is a,c and d only. please check

[@studywithmeguyz]

In 1st question :- a polynomial and its derivative is always continuous and bdd so it satisfies Lipschitz condition
Hence there is a unique solution in an interval containing 0 (d) is true

[@Zeelkumar31]

Option D is correct😊..!!

[@poojasingh7200]

​@@Zeelkumar31yeah D is also correct bcz
It's taking about A solution
That is one so

[@sibasankarbarada4169]

But sir e^z is unbounded because it is entire

[@naveenpondara4903]

No bro e^z is bounded

[@VlogswithJay]

My dear brother who told you that e^z is bounded its unbounded

[@naveenpondara4903]

@@VlogswithJay Haa yes...

[@VlogswithJay]

Bounded or unbounded ?

[@naveenpondara4903]

@@VlogswithJay if there is no condition for real and imaginary parts then it is unbounded

[@aweirdhimalayanbiker]

1st ka d hoga

[@alokkprajapati]

Thanks Sir ❤

[@spp626]

Thank you sir

[@Learnmathematics-k5o]

Thank you sir ❤

[@positivethoughts6419]

Thank you so much sir 😊

[@subho7125]

Thank u sir❤

[@professorragavan9783]

Thank you sir

[@Sudeeppradhan1999]

Thank you sir ❤❤❤